1 /*
2 * Copyright (c) 1992, 1993
3 * The Regents of the University of California. All rights reserved.
4 *
5 * This software was developed by the Computer Systems Engineering group
6 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
7 * contributed to Berkeley.
8 *
9 * All advertising materials mentioning features or use of this software
10 * must display the following acknowledgement:
11 * This product includes software developed by the University of
12 * California, Lawrence Berkeley Laboratory.
13 *
14 * %sccs.include.redist.c%
15 *
16 * @(#)fpu_sqrt.c 8.1 (Berkeley) 06/11/93
17 *
18 * from: $Header: fpu_sqrt.c,v 1.3 92/11/26 01:39:51 torek Exp $
19 */
20
21 /*
22 * Perform an FPU square root (return sqrt(x)).
23 */
24
25 #include <sys/types.h>
26
27 #include <machine/reg.h>
28
29 #include <sparc/fpu/fpu_arith.h>
30 #include <sparc/fpu/fpu_emu.h>
31
32 /*
33 * Our task is to calculate the square root of a floating point number x0.
34 * This number x normally has the form:
35 *
36 * exp
37 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer)
38 *
39 * This can be left as it stands, or the mantissa can be doubled and the
40 * exponent decremented:
41 *
42 * exp-1
43 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4)
44 *
45 * If the exponent `exp' is even, the square root of the number is best
46 * handled using the first form, and is by definition equal to:
47 *
48 * exp/2
49 * sqrt(x) = sqrt(mant) * 2
50 *
51 * If exp is odd, on the other hand, it is convenient to use the second
52 * form, giving:
53 *
54 * (exp-1)/2
55 * sqrt(x) = sqrt(2 * mant) * 2
56 *
57 * In the first case, we have
58 *
59 * 1 <= mant < 2
60 *
61 * and therefore
62 *
63 * sqrt(1) <= sqrt(mant) < sqrt(2)
64 *
65 * while in the second case we have
66 *
67 * 2 <= 2*mant < 4
68 *
69 * and therefore
70 *
71 * sqrt(2) <= sqrt(2*mant) < sqrt(4)
72 *
73 * so that in any case, we are sure that
74 *
75 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2
76 *
77 * or
78 *
79 * 1 <= sqrt(n * mant) < 2, n = 1 or 2.
80 *
81 * This root is therefore a properly formed mantissa for a floating
82 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2
83 * as above. This leaves us with the problem of finding the square root
84 * of a fixed-point number in the range [1..4).
85 *
86 * Though it may not be instantly obvious, the following square root
87 * algorithm works for any integer x of an even number of bits, provided
88 * that no overflows occur:
89 *
90 * let q = 0
91 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
92 * x *= 2 -- multiply by radix, for next digit
93 * if x >= 2q + 2^k then -- if adding 2^k does not
94 * x -= 2q + 2^k -- exceed the correct root,
95 * q += 2^k -- add 2^k and adjust x
96 * fi
97 * done
98 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x)
99 *
100 * If NBITS is odd (so that k is initially even), we can just add another
101 * zero bit at the top of x. Doing so means that q is not going to acquire
102 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the
103 * final value in x is not needed, or can be off by a factor of 2, this is
104 * equivalant to moving the `x *= 2' step to the bottom of the loop:
105 *
106 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
107 *
108 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
109 * (Since the algorithm is destructive on x, we will call x's initial
110 * value, for which q is some power of two times its square root, x0.)
111 *
112 * If we insert a loop invariant y = 2q, we can then rewrite this using
113 * C notation as:
114 *
115 * q = y = 0; x = x0;
116 * for (k = NBITS; --k >= 0;) {
117 * #if (NBITS is even)
118 * x *= 2;
119 * #endif
120 * t = y + (1 << k);
121 * if (x >= t) {
122 * x -= t;
123 * q += 1 << k;
124 * y += 1 << (k + 1);
125 * }
126 * #if (NBITS is odd)
127 * x *= 2;
128 * #endif
129 * }
130 *
131 * If x0 is fixed point, rather than an integer, we can simply alter the
132 * scale factor between q and sqrt(x0). As it happens, we can easily arrange
133 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
134 *
135 * In our case, however, x0 (and therefore x, y, q, and t) are multiword
136 * integers, which adds some complication. But note that q is built one
137 * bit at a time, from the top down, and is not used itself in the loop
138 * (we use 2q as held in y instead). This means we can build our answer
139 * in an integer, one word at a time, which saves a bit of work. Also,
140 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
141 * `new' bits in y and we can set them with an `or' operation rather than
142 * a full-blown multiword add.
143 *
144 * We are almost done, except for one snag. We must prove that none of our
145 * intermediate calculations can overflow. We know that x0 is in [1..4)
146 * and therefore the square root in q will be in [1..2), but what about x,
147 * y, and t?
148 *
149 * We know that y = 2q at the beginning of each loop. (The relation only
150 * fails temporarily while y and q are being updated.) Since q < 2, y < 4.
151 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
152 * Furthermore, we can prove with a bit of work that x never exceeds y by
153 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as
154 * an exercise to the reader, mostly because I have become tired of working
155 * on this comment.)
156 *
157 * If our floating point mantissas (which are of the form 1.frac) occupy
158 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
159 * In fact, we want even one more bit (for a carry, to avoid compares), or
160 * three extra. There is a comment in fpu_emu.h reminding maintainers of
161 * this, so we have some justification in assuming it.
162 */
163 struct fpn *
fpu_sqrt(fe)164 fpu_sqrt(fe)
165 struct fpemu *fe;
166 {
167 register struct fpn *x = &fe->fe_f1;
168 register u_int bit, q, tt;
169 register u_int x0, x1, x2, x3;
170 register u_int y0, y1, y2, y3;
171 register u_int d0, d1, d2, d3;
172 register int e;
173
174 /*
175 * Take care of special cases first. In order:
176 *
177 * sqrt(NaN) = NaN
178 * sqrt(+0) = +0
179 * sqrt(-0) = -0
180 * sqrt(x < 0) = NaN (including sqrt(-Inf))
181 * sqrt(+Inf) = +Inf
182 *
183 * Then all that remains are numbers with mantissas in [1..2).
184 */
185 if (ISNAN(x) || ISZERO(x))
186 return (x);
187 if (x->fp_sign)
188 return (fpu_newnan(fe));
189 if (ISINF(x))
190 return (x);
191
192 /*
193 * Calculate result exponent. As noted above, this may involve
194 * doubling the mantissa. We will also need to double x each
195 * time around the loop, so we define a macro for this here, and
196 * we break out the multiword mantissa.
197 */
198 #ifdef FPU_SHL1_BY_ADD
199 #define DOUBLE_X { \
200 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
201 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
202 }
203 #else
204 #define DOUBLE_X { \
205 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
206 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
207 }
208 #endif
209 #if (FP_NMANT & 1) != 0
210 # define ODD_DOUBLE DOUBLE_X
211 # define EVEN_DOUBLE /* nothing */
212 #else
213 # define ODD_DOUBLE /* nothing */
214 # define EVEN_DOUBLE DOUBLE_X
215 #endif
216 x0 = x->fp_mant[0];
217 x1 = x->fp_mant[1];
218 x2 = x->fp_mant[2];
219 x3 = x->fp_mant[3];
220 e = x->fp_exp;
221 if (e & 1) /* exponent is odd; use sqrt(2mant) */
222 DOUBLE_X;
223 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
224 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */
225
226 /*
227 * Now calculate the mantissa root. Since x is now in [1..4),
228 * we know that the first trip around the loop will definitely
229 * set the top bit in q, so we can do that manually and start
230 * the loop at the next bit down instead. We must be sure to
231 * double x correctly while doing the `known q=1.0'.
232 *
233 * We do this one mantissa-word at a time, as noted above, to
234 * save work. To avoid `(1 << 31) << 1', we also do the top bit
235 * outside of each per-word loop.
236 *
237 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
238 * t3 = y3, t? |= bit' for the appropriate word. Since the bit
239 * is always a `new' one, this means that three of the `t?'s are
240 * just the corresponding `y?'; we use `#define's here for this.
241 * The variable `tt' holds the actual `t?' variable.
242 */
243
244 /* calculate q0 */
245 #define t0 tt
246 bit = FP_1;
247 EVEN_DOUBLE;
248 /* if (x >= (t0 = y0 | bit)) { */ /* always true */
249 q = bit;
250 x0 -= bit;
251 y0 = bit << 1;
252 /* } */
253 ODD_DOUBLE;
254 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */
255 EVEN_DOUBLE;
256 t0 = y0 | bit; /* t = y + bit */
257 if (x0 >= t0) { /* if x >= t then */
258 x0 -= t0; /* x -= t */
259 q |= bit; /* q += bit */
260 y0 |= bit << 1; /* y += bit << 1 */
261 }
262 ODD_DOUBLE;
263 }
264 x->fp_mant[0] = q;
265 #undef t0
266
267 /* calculate q1. note (y0&1)==0. */
268 #define t0 y0
269 #define t1 tt
270 q = 0;
271 y1 = 0;
272 bit = 1 << 31;
273 EVEN_DOUBLE;
274 t1 = bit;
275 FPU_SUBS(d1, x1, t1);
276 FPU_SUBC(d0, x0, t0); /* d = x - t */
277 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */
278 x0 = d0, x1 = d1; /* x -= t */
279 q = bit; /* q += bit */
280 y0 |= 1; /* y += bit << 1 */
281 }
282 ODD_DOUBLE;
283 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */
284 EVEN_DOUBLE; /* as before */
285 t1 = y1 | bit;
286 FPU_SUBS(d1, x1, t1);
287 FPU_SUBC(d0, x0, t0);
288 if ((int)d0 >= 0) {
289 x0 = d0, x1 = d1;
290 q |= bit;
291 y1 |= bit << 1;
292 }
293 ODD_DOUBLE;
294 }
295 x->fp_mant[1] = q;
296 #undef t1
297
298 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */
299 #define t1 y1
300 #define t2 tt
301 q = 0;
302 y2 = 0;
303 bit = 1 << 31;
304 EVEN_DOUBLE;
305 t2 = bit;
306 FPU_SUBS(d2, x2, t2);
307 FPU_SUBCS(d1, x1, t1);
308 FPU_SUBC(d0, x0, t0);
309 if ((int)d0 >= 0) {
310 x0 = d0, x1 = d1, x2 = d2;
311 q |= bit;
312 y1 |= 1; /* now t1, y1 are set in concrete */
313 }
314 ODD_DOUBLE;
315 while ((bit >>= 1) != 0) {
316 EVEN_DOUBLE;
317 t2 = y2 | bit;
318 FPU_SUBS(d2, x2, t2);
319 FPU_SUBCS(d1, x1, t1);
320 FPU_SUBC(d0, x0, t0);
321 if ((int)d0 >= 0) {
322 x0 = d0, x1 = d1, x2 = d2;
323 q |= bit;
324 y2 |= bit << 1;
325 }
326 ODD_DOUBLE;
327 }
328 x->fp_mant[2] = q;
329 #undef t2
330
331 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */
332 #define t2 y2
333 #define t3 tt
334 q = 0;
335 y3 = 0;
336 bit = 1 << 31;
337 EVEN_DOUBLE;
338 t3 = bit;
339 FPU_SUBS(d3, x3, t3);
340 FPU_SUBCS(d2, x2, t2);
341 FPU_SUBCS(d1, x1, t1);
342 FPU_SUBC(d0, x0, t0);
343 ODD_DOUBLE;
344 if ((int)d0 >= 0) {
345 x0 = d0, x1 = d1, x2 = d2;
346 q |= bit;
347 y2 |= 1;
348 }
349 while ((bit >>= 1) != 0) {
350 EVEN_DOUBLE;
351 t3 = y3 | bit;
352 FPU_SUBS(d3, x3, t3);
353 FPU_SUBCS(d2, x2, t2);
354 FPU_SUBCS(d1, x1, t1);
355 FPU_SUBC(d0, x0, t0);
356 if ((int)d0 >= 0) {
357 x0 = d0, x1 = d1, x2 = d2;
358 q |= bit;
359 y3 |= bit << 1;
360 }
361 ODD_DOUBLE;
362 }
363 x->fp_mant[3] = q;
364
365 /*
366 * The result, which includes guard and round bits, is exact iff
367 * x is now zero; any nonzero bits in x represent sticky bits.
368 */
369 x->fp_sticky = x0 | x1 | x2 | x3;
370 return (x);
371 }
372